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Two taps can fill a cistern in 20 minutes and 30 minutes. The first

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Two taps can fill a cistern in 20 minutes and 30 minutes. The first  [#permalink]

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New post 01 Jul 2017, 03:36
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Two taps can fill a cistern in 20 minutes and 30 minutes. The first tap was opened initially for x minutes after which the second tap was opened. If it took a total of 15 minutes for the tank to be filled, what is the value of x?

(A) 5.0
(B) 7.5
(C) 9.0
(D) 10.0
(E) 12.5

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Re: Two taps can fill a cistern in 20 minutes and 30 minutes. The first  [#permalink]

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New post 01 Jul 2017, 08:31
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Bunuel wrote:
Two taps can fill a cistern in 20 minutes and 30 minutes. The first tap was opened initially for x minutes after which the second tap was opened. If it took a total of 15 minutes for the tank to be filled, what is the value of x?

(A) 5.0
(B) 7.5
(C) 9.0
(D) 10.0
(E) 12.5



hi,

a logical way to look at the Q is..
Tap I has worked for entire 15 minutes, so it fills up \(\frac{15}{20}=\frac{3}{4}\) of cistern..
Remaining is filled up by tap II..
so 1/4 will be filled by II in 30*1/4=7.5..
so x is 15-7.5=7.5
B
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Two taps can fill a cistern in 20 minutes and 30 minutes. The first  [#permalink]

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New post 01 Jul 2017, 03:45
2
1
rate of the 1st tap: 1/20 (per minute)
rate of the 2nd tap: 1/30 (per minute)

The first tap was opened initially for x minutes => Work done by the 1st tap in x minutes: x/20
After that, the second tap was opened. If it took a total of 15 minutes for the tank to be filled => Both taps were opened in (15-x) minutes => Work done by both taps during (15-x) minutes: (15-x)(1/20 + 1/30)
Therefore, we have this equation: x/20 + (15-x)(1/20 + 1/30) = 1
=> x=7.5

Option (B) is correct.
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Two taps can fill a cistern in 20 minutes and 30 minutes. The first  [#permalink]

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New post 01 Jul 2017, 08:25
Bunuel wrote:
Two taps can fill a cistern in 20 minutes and 30 minutes. The first tap was opened initially for x minutes after which the second tap was opened. If it took a total of 15 minutes for the tank to be filled, what is the value of x?

(A) 5.0
(B) 7.5
(C) 9.0
(D) 10.0
(E) 12.5

I backsolved here. Finding equation was taking too long.

1. Start with (C), where Tap 1 is open by itself for 9 minutes.

Tap 1 rate = \(\frac{1}{20}\)

W = rt, so \(\frac{1}{20}\) * 9 = \(\frac{9}{20}\) work is completed --->

\(\frac{11}{20}\) work remains.

9 minutes have passed.

2. Then Tap 1 and Tap 2 work together. Add rates.

\(\frac{1}{20}\) + \(\frac{1}{30}\) = \(\frac{50}{600}\) = \(\frac{1}{12}\)

W/r = t: \(\frac{11}{20}\) divided by \(\frac{1}{12}\) = \(\frac{132}{60}\) = 6.6 minutes

9 minutes elapsed + 6.6 more = 15.6 minutes. x = 9 is just a little too large. Try (B).

3. Answer (B), Tap 1 alone for 7.5 = \(\frac{15}{2}\) minutes

Work done by Tap 1: \(\frac{1}{20}\) * \(\frac{15}{2}\) = \(\frac{3}{8}\)

\(\frac{5}{8}\) of work remains. 7.5 minutes have elapsed.

4. Taps 1 and 2 work together. From above, combined rate is \(\frac{1}{12}\).

W/r = t: \(\frac{5}{8}\) divided by \(\frac{1}{12}\) =

\(\frac{15}{2}\) = 7.5 minutes.

7.5 minutes elapsed + 7.5 minutes more = 15 minutes. Answer B.
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Re: Two taps can fill a cistern in 20 minutes and 30 minutes. The first  [#permalink]

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New post 01 Jul 2017, 13:36
Assume volume of the tank to be 600 units.
First tap takes 20 minutes(Rate : 30 units/minute)
Second tap takes 30 minutes(Rate : 20 units/minute)

Given data : The first tap was opened initially for x minutes
Total time taken to fill the tank was 15 minutes
Since we know that the taps combined fill (20+30 units/minute = 50 units/minute)

From question stem, we can frame equation
30x + 50(15-x) = 600

Solving for x,
30x + 750 - 50x = 600
750 - 600 = 50x - 30x
150 = 20x
x = 7.5 minutes(Option B)

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Re: Two taps can fill a cistern in 20 minutes and 30 minutes. The first  [#permalink]

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New post 10 Jul 2017, 05:01
1
1
Bunuel wrote:
Two taps can fill a cistern in 20 minutes and 30 minutes. The first tap was opened initially for x minutes after which the second tap was opened. If it took a total of 15 minutes for the tank to be filled, what is the value of x?

(A) 5.0
(B) 7.5
(C) 9.0
(D) 10.0
(E) 12.5


Responding to a pm:

Another way to look at it:

Combined rate of both taps = 1/20 + 1/30 = 5/60 = 1/12

If both worked together, it would have taken 12 mins. 3 mins less will be taken if the second tap also works for the whole time. So in 3 mins, both taps do 1/4 of the work. That is what the second tap would have done extra (if it were open for the whole time)

Time in which second tap did 1/4 of the work = (1/4)/(1/30) = 7.5 mins

So the second tap was on for 7.5 mins and the first tap was on alone for 15-7.5 = 7.5 mins = x

Answer (B)
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Re: Two taps can fill a cistern in 20 minutes and 30 minutes. The first  [#permalink]

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New post 10 Jul 2017, 06:15
\(\frac{x}{20} + 15-x(\frac{1}{20}+\frac{1}{30}) = 1\)

\(\frac{x}{20} + \frac{15-x}{20} + \frac{15-x}{30} = 1\)

\(\frac{15}{20} + \frac{15-x}{30} = 1\)

\(450 + 300 - 20x = 600\)

\(20x = 150\)

\(x = 7.5\). Ans - B.
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Re: Two taps can fill a cistern in 20 minutes and 30 minutes. The first  [#permalink]

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New post 12 Jul 2017, 16:56
1
2
Bunuel wrote:
Two taps can fill a cistern in 20 minutes and 30 minutes. The first tap was opened initially for x minutes after which the second tap was opened. If it took a total of 15 minutes for the tank to be filled, what is the value of x?

(A) 5.0
(B) 7.5
(C) 9.0
(D) 10.0
(E) 12.5


We are given that the rate of the first cistern is 1/20 and the rate of the second cistern is 1/30. The first tap worked for 15 minutes and the second tap worked for (15 - x) minutes. Thus:

(1/20)(15) + (1/30)(15 - x) = 1

3/4 + (15 - x)/30 = 1

Multiplying the entire equation by 60, we have:

45 + 30 - 2x = 60

2x = 15

x = 7.5

Answer: B
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Re: Two taps can fill a cistern in 20 minutes and 30 minutes. The first  [#permalink]

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Re: Two taps can fill a cistern in 20 minutes and 30 minutes. The first &nbs [#permalink] 30 Jul 2018, 00:24
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